Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 11.5 ALTERNATING SERIES 0 65
10 cos 2 1 cos 2 2 cos 2 3 cos 2 10 cos 2 n 1 .
35. I; s-n cos 2 n = ~ + --+ --+ .. · + - - - ~ 0.07393. Now -- < - so the error IS
n =l 5 5 2 5 3 5 10 5" - 5n'
l
oo 1 . j·t -x . [ 5- "' ] ' . ( 5 - t 5-1o) 1 _8
R10 ~ T10 ~ -dx= lim 5 dx= lim - ln = lim - - 5 1 5
+ ln S = 510 ln < 6.4 x 10 .
10 5"' t~oo 10 t~oo 10 t~ oo n
5
d,. 9 d . ~ 9 . . . . (I I 1 1) o d d d ~ dn
37. Since - < - for each n, an smce L.., - 1s a convergent geometnc senes r = 10 < , . 1 2 3 . .. = L.., -
10" - 10" n=1 10n n = l 10"
will always converge by the Comparison Test.
39. Since I: a .. converges, lim an = 0, so there exists N such that ian - 01 < 1 for all n > N => . 0 ~ ar,. < 1 for
n->oo
all n > N => 0 ~ a~, ~ an. Since I; an converges, so does I; a~ by the Comparison Test.
41 . (a) S ince lim an = oo, there is an integer N such that abn > 1 whenever n > N. (Take M = 1 in Definition 11.1.5.)
-oo~ n .
Then an > bn whenever n > Nand since I: b,. is divergent, I; a" is also divergent by the Comparison Test.
. 1 d b 1 fi > 2 h I' an lim. n lim X II li 1 lim
(b) (1) If an = - an n = - or n _ . , t en 1m -b = - 1
. - = -ln = m - 1 = x = oo,
Inn n n-oo n n ........ oo n n X--+00 X x-oo 1 X x-oo
~ 1 . .
so by part (a), L.., . -ln IS d1vergent.
n=2 n
.. ln n d b
1 th ~ b · h d' h · · d I' an
n n n=l n--+oo n.
( 11) If an = -- an n = -, en L.., n 1s t e 1vergent armomc senes an 1m -b
lim Inn = lim lnx = oo,
,,_ oo x-oo
00
so I: an. diverges by part (a).
n=l
43. lim nan = lim a/,. , so we apply the Limit Comparison Test with bn = .! . Since lim nan > 0 we know that either both
n--+oo n - oo 1 n n ,,_00
series converge or both series diverge, and we also know that f .! diverges [p-serics with p = 1]. Therefore, I; an must be
. n =l n
divergent.
45. Yes. Since I: an is a convergent series with positive terms, lim an= 0 by Theorem 11.2.6, and I: bn =I: sin( an) is a
series with positive terms (for large enough n). We have lim bn = lim sin( an) = 1 > 0 by Tbeor~m 2.4.2
n-oo an n--+oo an
n~oo
[ET Theorem 3.3.2]. Thus, l::bn is also convergent by the Limit Comparison Test.
11.5 Alternating Series
1. (a) An alternating series is a series whose terms are alternately positive and negative.
(b) An alternating series f; an= f; ( -l)"- 1 bn, where bn =ian I, converges if O < b,.+l ~ b,. for all nand lim b,, = 0.
n=l n=l n-+oo
(This is the Alternating Series Test.)
(c) The error involved in using the partial sum s.,. as an approximation to the total sum s is the remainder R ,. = s - sn and the
size of the error is smaller than bn+l; that is, IR~,I ~ bn+l· (This is the Alternating Series Estimation Theor~m.)
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